Exist any common definition of simple objects in derived categories, or even better, in stable model categories? I was only able to find definition for abelian categories.
Thanks.
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$\begingroup$ You probably have something in mind which makes the obvious definition (has no proper subobjects) not a good choice? $\endgroup$– Mariano Suárez-ÁlvarezCommented Jul 11, 2011 at 17:05
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1$\begingroup$ This is probably me not knowing enough, but what is the notion of a subobject in a derived category. $\endgroup$– Karl SchwedeCommented Jul 11, 2011 at 17:29
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1$\begingroup$ You might want to find a notion of simple object in a derived category such that, if the category has t-structure, an object in the heart of the t-structure is simple if and only if it is simple in the abelian category sense. I don't believe that there is such a notion which is independent of the choice of t-structure, but I'm not sure... $\endgroup$– Sam GunninghamCommented Jul 11, 2011 at 19:30
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3$\begingroup$ Note that the naive definition using the general category theory notion of subobject (which I find quite unnatural to apply to derived/stable $\infty$-type categories) does not have this property. $\endgroup$– Sam GunninghamCommented Jul 11, 2011 at 19:32
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1$\begingroup$ What do you want to use the definition for? In the category of spectra would you like the Eilenberg spectrum $H\mathbb Z$ to be considered simple? $\endgroup$– Tom GoodwillieCommented Jul 11, 2011 at 21:42
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2 Answers
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Sometimes an object $E$ of a $k$-linear category $C$ is called simple if $Hom_C(E,E) = k$. This notion is frequently used in derived categories.
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8$\begingroup$ That's a strange definition. For lots of abelian categories, objects with that property are in fact not simple in the sense of not having proper subobjects. (Some) representation theorists say such an object is a brick. $\endgroup$ Commented Jul 11, 2011 at 23:09
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$\begingroup$ (and they tend to be indecomposable, not simple...) $\endgroup$ Commented Jul 11, 2011 at 23:11
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1$\begingroup$ I am not sure who is responsible for using the word ``simple'' for this property, but the fact is that this terminology is used in literature. $\endgroup$– SashaCommented Jul 12, 2011 at 4:50
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In an arbitrary category, you can define subobjects of $X$ to be equivalence classes of pairs $(A,f)$, where $f:A\to X$ is a monomorphism, and $(A,f)$ is equivalent to $(B,g)$ if there is an isomorphism $p:A\to B$ with $gp=f$. You can then say that an object in an additive is simple if it has only the two obvious subobjects (and they are different). In a triangulated category, any monomorphism $f:A\to X$ is split, because there is a distinguished triangle $W\xrightarrow{m}A\xrightarrow{f}X\xrightarrow{n}\Sigma W$, and $fm=0$ so $m=0$. This means that the simple objects are precisely those that are indecomposable under direct sums.
answered Jul 11, 2011 at 17:55
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$\begingroup$ Good explanation. It seems to me, this definition even works in every category with a zero object, not just abelian categories. But perhaps I'm missing something ? $\endgroup$– RalphCommented Jul 11, 2011 at 19:28
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4$\begingroup$ In my experience (coming mostly from the quiver theory literature), this definition isn't popular because there are very few monomorphisms in the derived category. For example, if $R$ is a ring, and $0 \to M_1 \to M_2 \to M_3 \to 0$ is a non-split s.e.s. of $R$-modules, then $M_1 \to M_2$ is not a monomorphism in the derived category of $R$-modules. In my reading, Sasha's definition is the standard one. $\endgroup$ Commented Jul 11, 2011 at 22:45
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2$\begingroup$ I may be missing something simple, but to justify your last sentence shouldn't we require that the triangulated category is Karoubian? $\endgroup$– NunoCommented Sep 26, 2015 at 17:46